# Some properties of meromorphically multivalent functions

## Abstract

By using the method of differential subordinations, we derive certain properties of meromorphically multivalent functions.

2010 Mathematics Subject Classification: 30C45; 30C55.

## 1 Introduction

Let Î£(p) denotes the class of meromorphically multivalent functions f(z) of the form

$f\left(z\right)={z}^{-p}+\underset{k=1}{\overset{\mathrm{âˆž}}{âˆ‘}}{a}_{k-p}{z}^{k-p}\phantom{\rule{1em}{0ex}}\left(pâˆˆN=\left\{1,2,3,â€¦\right\}\right),$
(1.1)

which are analytic in the punctured unit disk

${U}^{*}=\left\{z:zâˆˆC\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{and}}\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{0}}<\left|z\right|<1\right\}=U\\left\{0\right\}.$

Let f(z) and g(z) be analytic in U. Then, we say that f(z) is subordinate to g(z) in U, written f(z) â‰º g(z), if there exists an analytic function w(z) in U, such that |w(z)| â‰¤ |z| and f(z) = g(w(z)) (z âˆˆ U). If g(z) is univalent in U, then the subordination f(z) â‰º g(z) is equivalent to f(0) = g(0) and f(U) âŠ‚ g(U).

Let p(z) = 1 + p1z + ... be analytic in U. Then for -1 â‰¤ B < A â‰¤ 1, it is clear that

$p\left(z\right)â‰º\frac{1+Az}{1+Bz}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right)$
(1.2)

if and only if

$\left|p\left(z\right)-\frac{1-AB}{1-{B}^{2}}\right|<\frac{A-B}{1-{B}^{2}}\phantom{\rule{1em}{0ex}}\left(-1
(1.3)

and

$\mathsf{\text{Re}}p\left(z\right)>\frac{1-A}{2}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(B=-1;zâˆˆU\right).$
(1.4)

Recently, several authors (see, e.g., [1â€“7]) considered some interesting properties of meromorphically multivalent functions. In the present article, we aim at proving some subordination properties for the class Î£(p).

To derive our results, we need the following lemmas.

Lemma 1 (see [8]. Let h(z) be analytic and starlike univalent in U with h(0) = 0. If g(z) is analytic in U and zg'(z) â‰º h(z), then

$g\left(z\right)â‰ºg\left(0\right)+\underset{0}{\overset{z}{âˆ«}}\frac{h\left(t\right)}{t}dt.$

Lemma 2 (see [9]. Let p(z) be analytic and nonconstant in U with p(0) = 1. If 0 < | z0 | < 1 and $\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}p\left({z}_{0}\right)=\underset{\left|z\right|â‰¤\left|{z}_{0}\right|}{\text{min}}\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}p\left(z\right)$, then

${z}_{0}{p}^{\mathrm{â€²}}\left({z}_{0}\right)â‰¤-\frac{{\left|1-p\left({z}_{0}\right)\right|}^{2}}{2\left(1-\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}p\left({z}_{0}\right)\right)}.$

## 2 Main results

Our first result is contained in the following.

Theorem 1. Let $\mathrm{Î±}âˆˆ\left(0,\frac{1}{2}\right]$ and Î² âˆˆ (0,1). If f(z) âˆˆ Î£(p) satisfies f(z) â‰  0 (z âˆˆ U*) and

$\left|\frac{{z}^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+p\right)\right|<\mathrm{Î´}\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$
(2.1)

where Î´ is the minimum positive root of the equation

$\mathrm{Î±}\text{sin}\left(\frac{\mathrm{Ï€}\mathrm{Î²}}{2}\right){x}^{2}-x+\left(1-\mathrm{Î±}\right)\text{sin}\left(\frac{\mathrm{Ï€}\mathrm{Î²}}{2}\right)=0,$
(2.2)

then

$\left|\text{arg}\left(\frac{f\left(z\right)}{{z}^{-p}}-\mathrm{Î±}\right)\right|<\frac{\mathrm{Ï€}}{2}\mathrm{Î²}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(zâˆˆU\right).$
(2.3)

The bound Î² is the best possible for each $\mathrm{Î±}âˆˆ\left(0,\frac{1}{2}\right]$.

Proof. Let

$g\left(x\right)=\mathrm{Î±}\text{sin}\left(\frac{\mathrm{Ï€}\mathrm{Î²}}{2}\right){x}^{2}-x+\left(1-\mathrm{Î±}\right)\text{sin}\left(\frac{\mathrm{Ï€}\mathrm{Î²}}{2}\right).$
(2.4)

We can see that the Equation (2.2) has two positive roots. Since g(0) > 0 and g(1) < 0, we have

$0<\frac{\mathrm{Î±}}{1-\mathrm{Î±}}\mathrm{Î´}â‰¤\mathrm{Î´}<1.$
(2.5)

Put

$\frac{f\left(z\right)}{{z}^{-p}}=\mathrm{Î±}+\left(1-\mathrm{Î±}\right)p\left(z\right).$
(2.6)

Then from the assumption of the theorem, we see that p(z) is analytic in U with p(0) = 1 and Î± + (1 - Î±)p(z) â‰  0 for all z âˆˆ U. Taking the logarithmic differentiations in both sides of (2.6), we get

$\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+p=\frac{\left(1-\mathrm{Î±}\right)z{p}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Î±}+\left(1-\mathrm{Î±}\right)p\left(z\right)}$
(2.7)

and

$\frac{{z}^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+p\right)=\frac{\left(1-\mathrm{Î±}\right)z{p}^{\mathrm{â€²}}\left(z\right)}{{\left(\mathrm{Î±}+\left(1-\mathrm{Î±}\right)p\left(z\right)\right)}^{2}}$
(2.8)

for all z âˆˆ U. Thus the inequality (2.1) is equivalent to

$\frac{\left(1-\mathrm{Î±}\right)z{p}^{\mathrm{â€²}}\left(z\right)}{{\left(\mathrm{Î±}+\left(1-\mathrm{Î±}\right)p\left(z\right)\right)}^{2}}â‰º\mathrm{Î´}z.$
(2.9)

By using Lemma 1, (2.9) leads to

$\underset{0}{\overset{z}{âˆ«}}\frac{\left(1-\mathrm{Î±}\right){p}^{\mathrm{â€²}}\left(t\right)}{{\left(\mathrm{Î±}+\left(1-\mathrm{Î±}\right)p\left(t\right)\right)}^{2}}dtâ‰º\mathrm{Î´}z$

or to

$1-\frac{1}{\mathrm{Î±}+\left(1-a\right)p\left(z\right)}â‰º\mathrm{Î´}z.$
(2.10)

In view of (2.5), (2.10) can be written as

$p\left(z\right)â‰º\frac{1+\frac{\mathrm{Î±}}{1-\mathrm{Î±}}\mathrm{Î´}z}{1-\mathrm{Î´}z}.$
(2.11)

Now by taking $A=\frac{\mathrm{Î±}}{1-\mathrm{Î±}}\mathrm{Î´}$ and B = -Î´ in (1.2) and (1.3), we have

$\begin{array}{ll}\hfill \left|\text{arg}\left(\frac{f\left(z\right)}{{z}^{-p}}-\mathrm{Î±}\right)\right|& =\left|\text{arg}p\left(z\right)\right|\phantom{\rule{2em}{0ex}}\\ <\text{arcsin}\left(\frac{\mathrm{Î´}}{1-\mathrm{Î±}+\mathrm{Î±}{\mathrm{Î´}}^{2}}\right)\phantom{\rule{2em}{0ex}}\\ =\frac{\mathrm{Ï€}}{2}\mathrm{Î²}\phantom{\rule{2em}{0ex}}\end{array}$

for all z âˆˆ U because of g(Î´) = 0. This proves (2.3).

Next, we consider the function f(z) defined by

$f\left(z\right)=\frac{{z}^{-p}}{1-{\mathrm{Î´}}_{z}}\phantom{\rule{2.77695pt}{0ex}}\left(zâˆˆ{U}^{*}\right).$

It is easy to see that

$\left|\frac{{z}^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+p\right)\right|=\left|\mathrm{Î´}z\right|<\mathrm{Î´}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(zâˆˆU\right).$

Since

$\frac{f\left(z\right)}{{z}^{-p}}-\mathrm{Î±}=\left(1-\mathrm{Î±}\right)\frac{1+\frac{\mathrm{Î±}}{1-\mathrm{Î±}}\mathrm{Î´}z}{1-\mathrm{Î´}z},$

it follows from (1.3) that

$\underset{zâˆˆU}{\text{sup}}\left|\text{arg}\left(\frac{f\left(z\right)}{{z}^{-p}}-\mathrm{Î±}\right)\right|=\text{arcsin}\left(\frac{\mathrm{Î´}}{1-\mathrm{Î±}+\mathrm{Î±}{\mathrm{Î´}}^{2}}\right)=\frac{\mathrm{Ï€}}{2}\mathrm{Î²}.$

Hence, we conclude that the bound Î² is the best possible for each $\mathrm{Î±}âˆˆ\left(0,\frac{1}{2}\right]$.

Next, we derive the following.

Theorem 2. If f(z) âˆˆ Î£(p) satisfies f(z) â‰  0 (z âˆˆ U*) and

$\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}\left\{\frac{{z}^{-p}}{f\left(z\right)}\left(\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}+p\right)\right\}<\mathrm{Î³}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(zâˆˆU\right),$
(2.12)

where

$0<\mathrm{Î³}<\frac{1}{2\phantom{\rule{2.77695pt}{0ex}}\text{log}\phantom{\rule{2.77695pt}{0ex}}2},$
(2.13)

then

$\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}\frac{{z}^{-p}}{f\left(z\right)}>1-2\mathrm{Î³}\phantom{\rule{2.77695pt}{0ex}}\text{log}\phantom{\rule{2.77695pt}{0ex}}2\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$
(2.14)

The bound in (2.14) is sharp.

Proof. Let

$p\left(z\right)=\frac{f\left(z\right)}{{z}^{-p}}.$
(2.15)

Then p(z) is analytic in U with p(0) = 1 and p(z) â‰  0 for z âˆˆ U. In view of (2.15) and (2.12), we have

$1-\frac{z{p}^{\mathrm{â€²}}\left(z\right)}{\mathrm{Î³}{p}^{2}\left(z\right)}â‰º\frac{1+z}{1-z},$

i.e.,

$z{\left(\frac{1}{p\left(z\right)}\right)}^{\mathrm{â€²}}â‰º\frac{2\mathrm{Î³}z}{1-z}.$

Now by using Lemma 1, we obtain

$\frac{1}{p\left(z\right)}â‰º1-2\mathrm{Î³}\phantom{\rule{2.77695pt}{0ex}}\text{log}\left(1-z\right).$
(2.16)

Since the function 1 - 2Î³ log(1 - z) is convex univalent in U and

$\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}\left(1-2\mathrm{Î³}\phantom{\rule{2.77695pt}{0ex}}\text{log}\left(1-z\right)\right)>1-2\mathrm{Î³}\phantom{\rule{2.77695pt}{0ex}}\text{log}2\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right),$

from (2.16), we get the inequality (2.14).

To show that the bound in (2.14) cannot be increased, we consider

$f\left(z\right)=\frac{{z}^{-p}}{1-2\mathrm{Î³}\phantom{\rule{2.77695pt}{0ex}}\text{log}\left(1-z\right)}\phantom{\rule{1em}{0ex}}\left(zâˆˆ{U}^{*}\right).$

It is easy to verify that the function f(z) satisfies the inequality (2.12). On the other hand, we have

$\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}\frac{{z}^{-p}}{f\left(z\right)}â†’1-2\mathrm{Î³}\text{log}2$

as z â†’ -1. Now the proof of the theorem is complete.

Finally, we discuss the following theorem.

Theorem 3. Let f(z) âˆˆ Î£(p) with f(z) â‰  0 (z âˆˆ U*). If

$\left|\mathsf{\text{Im}}\phantom{\rule{1em}{0ex}}\left\{\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}\left(\frac{f\left(z\right)}{{z}^{-p}}-Î»\right)\right\}\right|<\sqrt{Î»\left(Î»+2p\right)}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(zâˆˆU\right)$
(2.17)

for some Î»(Î» > 0), then

$\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}\frac{f\left(z\right)}{{z}^{-p}}>0\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(zâˆˆU\right).$
(2.18)

Proof. Let us define the analytic function p(z) in U by

$\frac{f\left(z\right)}{{z}^{-p}}=p\left(z\right).$

Then p(0) = 1, p(z) â‰  0 (z âˆˆ U) and

$\frac{z{f}^{\mathrm{â€²}}\left(z\right)}{f\left(z\right)}\left(\frac{f\left(z\right)}{{z}^{-p}}-Î»\right)=\left(p\left(z\right)-Î»\right)\left(\frac{z{p}^{\mathrm{â€²}}\left(z\right)}{p\left(z\right)}-p\right)\phantom{\rule{1em}{0ex}}\left(zâˆˆU\right).$
(2.19)

Suppose that there exists a point z0(0 < | z0 | < 1) such that

$\mathsf{\text{Re}}\phantom{\rule{1em}{0ex}}p\left(z\right)>0\phantom{\rule{2.77695pt}{0ex}}\left(\left|z\right|<\left|{z}_{0}\right|\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}p\left({z}_{0}\right)=i\mathrm{Î²},$
(2.20)

where Î² is real and Î² â‰  0. Then, applying Lemma 2, we get

${z}_{0}{p}^{\mathrm{â€²}}\left({z}_{0}\right)â‰¤-\frac{1+{\mathrm{Î²}}^{2}}{2}.$
(2.21)

Thus it follows from (2.19), (2.20), and (2.21) that

${I}_{0}=\mathsf{\text{Im}}\phantom{\rule{1em}{0ex}}\left\{\frac{{z}_{0}{f}^{\mathrm{â€²}}\left({z}_{0}\right)}{f\left({z}_{0}\right)}\left(\frac{f\left({z}_{0}\right)}{{z}_{0}^{-p}}-Î»\right)\right\}=-p\mathrm{Î²}+\frac{Î»}{\mathrm{Î²}}{z}_{0}{p}^{\mathrm{â€²}}\left({z}_{0}\right).$
(2.22)

In view of Î» > 0, from (2.21) and (2.22) we obtain

${I}_{0}â‰¥-\frac{Î»+\left(Î»+2p\right){\mathrm{Î²}}^{2}}{2\mathrm{Î²}}â‰¥\sqrt{Î»\left(Î»+2p\right)}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\left(\mathrm{Î²}<0\right)$
(2.23)

and

${I}_{0}â‰¤-\frac{Î»+\left(Î»+2p\right){\mathrm{Î²}}^{2}}{2\mathrm{Î²}}â‰¤-\sqrt{Î»\left(Î»+2p\right)}\phantom{\rule{2.77695pt}{0ex}}\left(\mathrm{Î²}>0\right).$
(2.24)

But both (2.23) and (2.24) contradict the assumption (2.17). Therefore, we have Rep(z) > 0 for all z âˆˆ U. This shows that (2.18) holds true.

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Correspondence to Jin-Lin Liu.

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Xu, YH., Yang, Q. & Liu, JL. Some properties of meromorphically multivalent functions. J Inequal Appl 2012, 86 (2012). https://doi.org/10.1186/1029-242X-2012-86